Two type of balls in a bag
In a bag there are $15$ red and $5$ white balls. Two balls are chosen at random and one is found to be red. The probability that the second one is also red is?
I have attempted this question by counting all the favorable cases:
- Both red $(15×14)$
- One red one white $(15×5)$
Our case is both red. The probability is, by Baye's theorem, $dfrac{15×14}{15×14+15×5}$. However, the answer is not $dfrac{14}{19}$ but $dfrac{7}{12}$.
probability
edited Nov 24 at 4:36
Saad
19.7k92252
asked Nov 24 at 3:56
CaptainQuestion
1165
add a comment |
In a bag there are $15$ red and $5$ white balls. Two balls are chosen at random and one is found to be red. The probability that the second one is also red is?
I have attempted this question by counting all the favorable cases:
- Both red $(15×14)$
- One red one white $(15×5)$
Our case is both red. The probability is, by Baye's theorem, $dfrac{15×14}{15×14+15×5}$. However, the answer is not $dfrac{14}{19}$ but $dfrac{7}{12}$.
probability
edited Nov 24 at 4:36
Saad
19.7k92252
asked Nov 24 at 3:56
CaptainQuestion
1165
Well done for sharing an attempt :)
– Shaun
Nov 24 at 4:07
add a comment |
In a bag there are $15$ red and $5$ white balls. Two balls are chosen at random and one is found to be red. The probability that the second one is also red is?
I have attempted this question by counting all the favorable cases:
- Both red $(15×14)$
- One red one white $(15×5)$
Our case is both red. The probability is, by Baye's theorem, $dfrac{15×14}{15×14+15×5}$. However, the answer is not $dfrac{14}{19}$ but $dfrac{7}{12}$.
probability
edited Nov 24 at 4:36
Saad
19.7k92252
asked Nov 24 at 3:56
CaptainQuestion
1165
In a bag there are $15$ red and $5$ white balls. Two balls are chosen at random and one is found to be red. The probability that the second one is also red is?
I have attempted this question by counting all the favorable cases:
- Both red $(15×14)$
- One red one white $(15×5)$
Our case is both red. The probability is, by Baye's theorem, $dfrac{15×14}{15×14+15×5}$. However, the answer is not $dfrac{14}{19}$ but $dfrac{7}{12}$.
probability
probability
edited Nov 24 at 4:36
Saad
19.7k92252
asked Nov 24 at 3:56
CaptainQuestion
1165
edited Nov 24 at 4:36
Saad
19.7k92252
asked Nov 24 at 3:56
CaptainQuestion
1165
edited Nov 24 at 4:36
Saad
19.7k92252
edited Nov 24 at 4:36
Saad
19.7k92252
edited Nov 24 at 4:36
Saad
19.7k92252
19.7k92252
asked Nov 24 at 3:56
CaptainQuestion
1165
asked Nov 24 at 3:56
CaptainQuestion
1165
asked Nov 24 at 3:56
CaptainQuestion
1165
1165
Well done for sharing an attempt :)
– Shaun
Nov 24 at 4:07
add a comment |
Well done for sharing an attempt :)
– Shaun
Nov 24 at 4:07
Well done for sharing an attempt :)
– Shaun
Nov 24 at 4:07
Well done for sharing an attempt :)
– Shaun
Nov 24 at 4:07
add a comment |
2 Answers
2
active
oldest
votes
You need to include the cases that you choose one white one red, which is the same as $(2)$. Thus, the probability is $$frac{15cdot14}{15cdot14+2cdot15cdot5}=frac7{12}.$$
answered Nov 24 at 4:14
Abraham Zhang
596312
add a comment |
In the second case (one red and one white) the count is not correct, since it does not account for the order of the balls. It should be first red and second white, plus first white and second red, in total $15cdot 5 + 5cdot 15 = 150$. So we have $15cdot 4 + 15cdot 5 + 5cdot 15 = 360$ pairs with at least one red ball, and $15*14 = 210$ pairs in which bot are red, hence $P = 210/360 = 7/12$.
answered Nov 24 at 4:14
mlerma54
1,087138
add a comment |
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2 Answers
2
active
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votes
2 Answers
2
active
oldest
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active
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active
oldest
votes
You need to include the cases that you choose one white one red, which is the same as $(2)$. Thus, the probability is $$frac{15cdot14}{15cdot14+2cdot15cdot5}=frac7{12}.$$
answered Nov 24 at 4:14
Abraham Zhang
596312
add a comment |
You need to include the cases that you choose one white one red, which is the same as $(2)$. Thus, the probability is $$frac{15cdot14}{15cdot14+2cdot15cdot5}=frac7{12}.$$
answered Nov 24 at 4:14
Abraham Zhang
596312
add a comment |
You need to include the cases that you choose one white one red, which is the same as $(2)$. Thus, the probability is $$frac{15cdot14}{15cdot14+2cdot15cdot5}=frac7{12}.$$
answered Nov 24 at 4:14
Abraham Zhang
596312
You need to include the cases that you choose one white one red, which is the same as $(2)$. Thus, the probability is $$frac{15cdot14}{15cdot14+2cdot15cdot5}=frac7{12}.$$
answered Nov 24 at 4:14
Abraham Zhang
596312
answered Nov 24 at 4:14
Abraham Zhang
596312
answered Nov 24 at 4:14
Abraham Zhang
596312
answered Nov 24 at 4:14
Abraham Zhang
596312
596312
add a comment |
add a comment |
In the second case (one red and one white) the count is not correct, since it does not account for the order of the balls. It should be first red and second white, plus first white and second red, in total $15cdot 5 + 5cdot 15 = 150$. So we have $15cdot 4 + 15cdot 5 + 5cdot 15 = 360$ pairs with at least one red ball, and $15*14 = 210$ pairs in which bot are red, hence $P = 210/360 = 7/12$.
answered Nov 24 at 4:14
mlerma54
1,087138
add a comment |
In the second case (one red and one white) the count is not correct, since it does not account for the order of the balls. It should be first red and second white, plus first white and second red, in total $15cdot 5 + 5cdot 15 = 150$. So we have $15cdot 4 + 15cdot 5 + 5cdot 15 = 360$ pairs with at least one red ball, and $15*14 = 210$ pairs in which bot are red, hence $P = 210/360 = 7/12$.
answered Nov 24 at 4:14
mlerma54
1,087138
add a comment |
In the second case (one red and one white) the count is not correct, since it does not account for the order of the balls. It should be first red and second white, plus first white and second red, in total $15cdot 5 + 5cdot 15 = 150$. So we have $15cdot 4 + 15cdot 5 + 5cdot 15 = 360$ pairs with at least one red ball, and $15*14 = 210$ pairs in which bot are red, hence $P = 210/360 = 7/12$.
answered Nov 24 at 4:14
mlerma54
1,087138
In the second case (one red and one white) the count is not correct, since it does not account for the order of the balls. It should be first red and second white, plus first white and second red, in total $15cdot 5 + 5cdot 15 = 150$. So we have $15cdot 4 + 15cdot 5 + 5cdot 15 = 360$ pairs with at least one red ball, and $15*14 = 210$ pairs in which bot are red, hence $P = 210/360 = 7/12$.
answered Nov 24 at 4:14
mlerma54
1,087138
answered Nov 24 at 4:14
mlerma54
1,087138
answered Nov 24 at 4:14
mlerma54
1,087138
answered Nov 24 at 4:14
mlerma54
1,087138
1,087138
add a comment |
add a comment |
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Well done for sharing an attempt :)
– Shaun
Nov 24 at 4:07